The smallest side of a right-angled triangle | vedclass

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(C) Step $1$: Find the side of the square.Perimeter of the square $= 56\, cm$.Side of the square $= \frac{56}{4} = 14\, cm$.Step $2$: Find the smalles

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$10$

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(C) Step $1$: Find the side of the square.Perimeter of the square $= 56\, cm$.Side of the square $= \frac{56}{4} = 14\, cm$.Step $2$: Find the smallest side of the right-angled triangle.Smallest side $= 14 - 8 = 6\, cm$.Step $3$: Find the length of the rectangle.Area of the rectangle $= 96\, cm^2$ and breadth $= 8\, cm$.Length $= \frac{96}{8} = 12\, cm$.Step $4$: Find the second largest side of the right-angled triangle.Second largest side $= 12 - 4 = 8\, cm$.Step $5$: Find the largest side (hypotenuse) of the right-angled triangle.Using the Pythagorean theorem,$c = \sqrt{a^2 + b^2}$.Largest side $= \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10\, cm$.

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